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GATE Mechanical Engineering notes on industrial engineering


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Read these free GATE Mechanical Engineering notes on Industrial Engineering. For full course, visit or call 9779434433. These notes are helpful in GATE Mechanical and other competitive engineering exams.

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GATE Mechanical Engineering notes on industrial engineering

  1. 1. 1 | 1. Industrial Engineering Introduction As per American Institute of Industrial Engineers (AIIE), industrial engineering is defined as follows: β€˜It is a branch of engineering concerned with the design, improvement and installation of integrated systems of people, materials, equipment and energy. β€˜ Industrial engineering is an engineering approach to the detailed analysis of the use and cost of the resources of an organisation. The main target for an industrial engineer is to achieve productivity improvement. Productivity improvement implies: More efficient use of resources Less waste per unit of input applied Higher levels of output for a fixed level of input Production Management Production management focuses on two major areas; 1. Design of the production system which includes product, process, plant, equipment, and 2. Development of the control systems to manage inventories, product quality, production schedules and productivity Factors related to design in PM cycle are: 1. Product design 2. Job and process design 3. Labour skills and training programs 4. Equipment selection 5. Material selection input 6. Plant selection and layout 7. Scheduling steps of the plan 8. Implementing and controlling the schedule 9. Operating the production system Factors related to control systems in PM cycle are: 1. Inventory control policies 2. Quality control policies 3. Production schedule control policies 4. Productivity and cost control policies 5. Constructing control systems 6. Implementing and operating control systems 7. Modifying policies and designs Production Management v/s Industrial Engineering Production management familiarizes a person with concepts and techniques specific to the analysis and management of a production activity Industrial engineering deals with the analysis, design and control of productive systems, i.e. the system produces either a product or a service E.g. The training of an aircraft pilot is analogous to management education, whereas the designing of the aircraft is analogous to industrial engineering education. It is assumed that industrial engineers do not operate the systems they design. Managerial Economics Principles:- There are four economic principles that managers should keep in mind; 1. The incremental principle – the decision is considered good if it increases revenue as compared to costs 2. The principle of time perspective – the decision should take into account the long term and short term effects on costs and revenue 3. The Opportunity cost principle – decision making should carefully measure the sacrifices required by the various alternatives 4. The discounting principle – if a decision affects the costs and revenues of a future date, it is necessary to discount these costs and revenues to present values Concept of Production Production is any process developed to transform a set of input elements like men, materials, capitals, information and energy into a specified set of output elements like finished products and services in proper quantity and quality. Factors of production are as follows: 1. Nature (land and other natural resources) 2. Labour (human efforts) 3. Capital (factory building, machinery, tools, raw materials etc.) 4. Enterprise (activity that organises the factors of production into an operating unit) Productivity It is analogous to the efficiency of a machine. Like we desire to increase the efficiency of a machine, it is desired to increase the productivity within the available resources. Mathematically it is defined as the ratio of output and input. Example: Plant A Plant B No. of workers 200 300 No. of items produced per unit time 10 20 Productivity 10 200 = 1 20 20 300 = 1 15 Factors affecting productivity: 1. Human resources 2. Technology and Capital Investment 3. Product or system design 4. Machinery and equipment
  2. 2. 2 | 5. Skill and effectiveness of the worker 6. Production volume Increasing the productivity of resources: This implies, producing more number of goods from the same amount of input (or resources) Plant location A plant is a place, where men, materials, money, equipment, machinery are brought together for manufacturing products. Factors governing plant location: Nearness to Raw material (reduces transportation cost) Transport facilities (like road, rail etc.) Nearness to markets (reduces transportation cost and the danger of damage to the finished product before reaching the customer) Availability of Labour (supply of stable and trained work force) Availability of water resources (for industries such as paper and chemical industries) Climatic conditions Financial and other aids Land (topography, area, shape, cost, drainage etc. influence the plant location choice) Plant Layout Plant layout means the disposition of facilities like equipments, material, manpower etc. and the services of the plant within the area of the site selected for plant setup. It begins with the design of the factory building and extends up to the location and movement of a work table. All the equipments and resources are given a proper place. Objectives of a good plant layout: 1. Material handling and transportation is minimized and efficiently controlled 2. Bottlenecks and points of congestion are eliminated (to make the raw material and semi finished goods move fast between consecutive work stations) 3. Suitable and adequate spaces are allocated to production centres and service centres 4. Minimizing the workers movement 5. Enhancing safety of the working conditions for all employees 6. Increased flexibility in design changes for future changes and expansion 7. Reduced plant maintenance cost Principles of Plant layout: a) Integration of production facilities in an efficient manner b) Minimum movements and material handling c) Smooth and continuous flow, by implementing proper line balancing techniques d) Cubic space utilization by saving the floor space for storage and making use of ceiling e) Safe and improved environments in shape of safe and efficient work places f) Flexibility for accommodating changing product designs and production process Process Layout It is also known as functional layout and is characterized by keeping similar machines or similar operations at one location. In simpler words, all lathes will be in one place, all milling machines at another and so on. This type of layout is used in industries engaged in non repetitive type of maintenance or manufacturing activities. Advantages of process layout: 1. Greater flexibility in regards to allotment of work to workers and equipment 2. Better and efficient utilization of available equipment 3. Reduced capital investment, as the number of machines or equipments required are less 4. Better product quality obtained 5. Workers in one section is not affected by the operation carried out in another section Disadvantages of Process Layout 1. More space is needed 2. Automatic material handling is a difficult task 3. More material-in-process remains in queue for subsequent operations 4. Production control is difficult 5. Requires more inspections and constant efficient co-ordination Product Layout It is also known as line layout, which means that various operations on the raw material is performed in a sequence and the machines are placed along the product flow line. This type of layout is preferred for industries where continuous production is performed. Advantages of product layout: 1. Less space requirements 2. Automatic material handling is easier 3. Material movement and handling, time and costs are less 4. Less in-process inventory 5. Product completion in less time
  3. 3. 3 | 6. Smooth and continuous work flow 7. Less skilled workers may serve the purpose Disadvantages of product layout: 1. Layout flexibility is considerably reduced 2. The pace of the process depends upon the output rate of the slowest machine. This increases the idle time 3. More number of machines of a particular have to be purchased in order to create adequate number of standbys in case of any failure. This increases the capital investment 4. It is very difficult to increase the capacity of the production beyond the layout capacity Combination Layout This type of layout combines the advantages of both, process and product layout. These kinds of layouts are very rare. This kind of a layout is possible where an item is being made in different types and sizes. In these kind of cases, the machinery is arranged in a process layout but the process grouping is then arranged in a sequence to manufacture various size and types of products. No matter the product varies in size and type, the operation sequence remain the same. Fixed Position Layout This kind of layout is inherent in ship building, aircraft building and big pressure vessel fabrication. In this type of layout, the men, materials and equipment move past the stationary product. Advantages of Position layout: 1. One or more skilled workers can be employed from the beginning till the end of the job to ensure continuity of the process 2. It involves least movement of materials 3. Maximum flexibility available for products and process 4. Different projects can be taken up for the same layout Disadvantages of Position layout: 1. Low content of work-in progress 2. Low utilization of labour and equipment 3. Involves high equipment handling costs Flow Pattern Achieving an optimum efficient flow of materials is one of the most important phases in plant layout. The principle for optimum effective flow is minimum movement. The principle of minimum movement reduces material handling costs, in-process inventory and space for processing. A flow pattern must be simple in order to have an easy supervision and control Various flow patterns along with their characteristics are given below in the form of a table; Line flow It is the simplest, the material enters at one end (X) and leaves at the other end (Y). Used in buildings having long lengths and smaller widths. L type flow It resembles Line flow, but is used in buildings where width is more as compared to line flow type buildings Circular flow It is preferred for rotary handling systems. Different work stations are located along the circular path. Raw material enters at X and finished product leaves at Y. U type flow In this the supervision is simpler as compared to Line flow and L type flow. The raw material and finished product from the same side. Preferred in square shaped buildings. Combination of line flow and circular type As compared to line flow, this system needs smaller building lengths. Processing upwards In this the material is processed while moving upwards or downwards in a multi storeyed
  4. 4. 4 | Work Station Design The work station design affects the production rates, efficiency and the accuracy with which an operation can be performed. Line Balancing & Process Planning Line Balancing Line balancing means balancing the line, i.e. balancing a production line or an assembly line. Let us consider three machines, A, B and C, which can process 5, 10 and 15 pieces per unit time respectively. There is a precedence constraint of A to B to C. Since machine A has the minimum capacity, this will make machine B idle for 50% of the time, and machine C idle for 66.66% of the time. This indicates that the line is unbalanced. One method to balance this line is to have 3 machines of type A, 2 machines of type B and one machine of type C. Another method to make sure that the machines B and C do not remain idle is to give some additional work to them. The main task of line balancing is to ensure that the tasks are evenly distributed among men, machinery and thereby ensuring minimum idle time. Line balancing aims at grouping the facilities and tasks and workers in an efficient pattern in order to obtain an optimum balance of the capacities of the processes. The tasks are grouped in such a way so that their total time is preferably equal to or a little lesser than the time available at each work station. Methods for Line Balancing 1. Heuristic method 2. Linear Programming Model 3. Dynamic Programming 4. Comsoal (a computer method for assembly line sequencing) For intermittent flow pattern, Heuristic method is used as they are simple and involve less time and money. For continuous flow pattern involving high volume production we choose between linear programming and dynamic programming. Heuristic Method: In this method a precedence diagram is drawn in a particular way which indicates the flexibility available for transferring tasks laterally from one column to another. The following procedural steps are taken in this method: 1. Identify the work 2. Break down the work into elemental tasks or steps 3. List the various steps as shown below; Steps or elemental tasks Immediate Predecessor Duration of the task (minutes) 1 - 3 2 - 4 3 1 2 4 2 5 5 3 4 6 4 8 7 5 2 8 7 4 9 8 6 (Total time=38 minutes) 4. Sketch the precedence diagram and mark the task duration 5. Assume that the maximum time available for this problem at any work station is 10 minutes. Or in other words, cycle time is 10 minutes. The total duration of tasks is 38 minutes which means that the minimum number of work stations required are 38/10 = 4. The maximum number of stations may be equal to as many as the number of tasks or steps, i.e. nine. 6. Two basic concepts of assigning tasks to stations: a) Permutability of tasks: It means that any number of tasks of a column can be combined to make up their total time closer to cycle time, provided their total time does not exceed the cycle time. Analysis is carried out column by column and one can move to next column only after the tasks in the previous column have been assigned to a station. b) Lateral transferability of the tasks: For making total time of tasks equal to cycle time, tasks or steps may be shifted laterally provided the precedence relationships are maintained. Keeping these two concepts in mind, the above given figure (precedence diagram) is modified to the figure given below. In the below given figure, tasks 1, 2 and 3 are grouped together to station A. Task 4 has been laterally shifted from column II to column III and has been grouped with task 5, occupying station B. similarly tasks 6 and 7, 8 and 9 have been grouped and placed at stations C and D, respectively. This way all the nine steps have distributed to four stations.
  5. 5. 5 | Linear Programming method of Line Balancing: Assume that a job is broken down into 6 elemental tasks and the total duration of all such tasks is 28 minutes. The cycle time at each work station is 10 minutes. Thus the minimum number of work stations required are 28/10=3 and the maximum number of work stations may be 6, i.e. equal to the number of tasks involved. The problem now reduces to find out the exact number of work stations needed and which tasks will be assigned to which station, as shown in the precedence diagram shown below. Process Planning A process is defined as any group of actions performed to achieve some output from an operation in accordance with a specified measure of effectiveness. During designing a product, some specifications are established; like physical dimensions, tolerances, standards and quality. The decision of specific details on achieving the desired outcome is a part of Process Planning. Process Planning is the systematic determination of the methods by which a product is to be manufactured, economically and competitively. It is the intermediate step between designing a product and manufacturing it. Process planning takes as its inputs the drawings or other specifications which indicate what is to be made and also the forecasts, orders or contracts which indicate how many are to be made. Information required to do Process Planning 1. Quantity of work to be done along with specifications 2. Quality of work to be completed 3. Availability of equipments, tools and personnel 4. Sequence of operations on raw material 5. Names of equipments on which operations will be performed 6. Standard time for each operation 7. When the operations will be performed Process Planning Procedure 1. Preparation of working drawings 2. Deciding to make or buy 3. Selection of manufacturing process 4. Machine capacity and equipment selection 5. Material selection and bill of material 6. Selection of jigs, fixtures and other attachments 7. Operation planning and tooling requirements 8. Preparation of documents such as operation and route sheets Example: ABC Company manufactures and sells gas stoves. It makes some of the parts for the gas stoves and purchases the rest. The engineering department believes that it might be possible to cut costs by manufacturing one of the parts currently being purchased for Rs 8.50 each. The firm uses 1,00,000 of these parts every year, and the accounting department compiles the following list of costs based on engineering estimates. Fixed costs will increase by Rs 50,000 Labour costs will increase by Rs 1,25,000 Factory overheads, currently running Rs 5,00,000 p.a. may increase by 12% Raw materials used to make the part will cost Rs 6,00,000 Given the above estimates, should ABC Company make the part or buy it? Solution: Calculate the total part incurred if the part was manufactured: Additional fixed costs Rs 50,000 Additional labour costs Rs 1,25,000 Raw materials costs Rs 6,00,000 Additional overheads costs Rs 60,000 ( 0.12 X 5,00,000) Total cost to manufacture Rs 8,35,000 Cost to manufacture one part Rs 8,35,000/ 1,00,000 = Rs 8.35 As compared to buying cost per piece, the cost of manufacturing one piece is less, therefore, the company should manufacture the part instead of buying it.
  6. 6. 6 | Process Analysis It means the study of the overall process in a plant. It analyses each step of the manufacturing process and aims at improving the industrial operations. It helps in finding better methods of doing a job, and this is achieved by eliminating the unproductive and unnecessary elements of the process or through modified layout of facilities. A process is analysed with the help of Process Charts and Flow diagrams. The steps involved in Process Analysis are: 1. Select the process for analysis 2. Break down the process into operations and sub-operations 3. Construct a process chart and a flow diagram 4. Analyse the process chart and flow diagram by subjecting each and every step to detailed questioning 5. Reconstruct the process chart and flow diagram 6. Test the proposed method for all advantages claimed 7. Explain the new method to the workers before putting it into place Process Chart It is a diagram which gives an overall view of the situation, in this case a process. It helps visualising various possibilities of improvement. A chart representing a process may be called a Process Chart. The chart illustrates the process with the help of a set of symbols and helps in better understanding. Process Chart Symbols Event Symbol Operation Storage Temporary Storage Transport Inspection Operation cum transportation Inspection cum operation Manual Process Planning This type of planning is called man-variant process planning and is the commonest type of planning used for production today The planner selects the combination of processes required to produce a finished part. In selecting this combination of processes, a number of criteria are employed Production cost or time are the dominant criteria in process selection; but machine utilization and routing often affects the plans chosen Automated Process Planning Man variant planning is often a boring and tedious job, with errors in human judgement To eliminate this automated process planning has come to aid of industrial process planning Figure (a) below shows a completely automated process planning system Figure (b) shows a fully automated system with human assistance to code the engineering drawing data
  7. 7. 7 | Production planning and control (PPC) Introduction Production is done my manufacturing different things with various processes. Planning looks ahead, anticipates possible difficulties and decides in advance about the production. The control phase makes sure that the programmed production is constantly maintained. A production planning and control (PPC) system has many functions to perform like:- Planning phase:- Forecasting, order writing, product, product design, material control, tool control, loading etc. Action phase:- Dispatching Control phase:- Data processing, expediting and replanning Continuous and Intermittent Production In continuous production, there is a continuous flow of material, which is achieved by using special machines and produces standardized items in large quantities. A continuous production system can be divided into two categories:- 1. Mass and flow line production 2. Continuous or process production In an intermittent production, there is an intermittent or interrupted flow of material. In this system we make use of general purpose machines and produce different components of different nature in small quantities. Intermittent production systems can be classified as:- 1. Batch production 2. Job production Job Shop, Open Job Shop and Closed Job Shop In a job shop, there is involvement of intermittent production. It consists of a number of machine centres, but each with a different activity to perform. In a job shop the material in-process follows different processing patterns in batches through batch facilities The material does not flow in a serial fashion A job shop makes to order and are not open to orders from just any source A closed job shop is one which is closed to job orders from outside the organization, e.g. machine shop of a big concern making automobile parts. In this standardized parts are made which have a certain market demand. An open job shop produces to order with a non-repetitive trend. In this kind of setup, the products are made to order as per the requirements of the customer. Forecasting It means estimation of type, quantity and quality of future work, e.g. sales etc. Forecasting plays an important role in planning for the future. The purpose of forecasting is:- To determine the production volume and rate To prepare production budget for production processes etc. Basic elements of forecasting are:- Trends Cycles Seasonal variations Irregular variations Forecasting techniques:- Historic estimate Sales force estimate Market survey Delphi method Judgemental techniques Prior knowledge Forecasting by past average Forecasting from last period sales Moving average method Weighted moving average method Exponential smoothening method Correlation analysis Regression analysis Forecasting by past average It is used when our aim is to forecast or predict the sales of an item for the next sales period. π‘’π‘ π‘‘π‘–π‘šπ‘Žπ‘‘π‘’π‘‘ π‘ π‘Žπ‘™π‘’π‘  π‘“π‘œπ‘Ÿ π‘‘β„Žπ‘’ 𝑛𝑒π‘₯𝑑 π‘π‘’π‘Ÿπ‘–π‘œπ‘‘ = π‘Žπ‘£π‘’π‘Ÿπ‘Žπ‘”π‘’ π‘ π‘Žπ‘™π‘’π‘  π‘“π‘œπ‘Ÿ π‘π‘Ÿπ‘’π‘£π‘–π‘œπ‘’π‘  π‘π‘’π‘Ÿπ‘–π‘œπ‘‘ Example: Period No. Sales 1 7 2 5 3 9 4 8 5 5 6 8 Forecasted sales for period number 7 = 7+5=9+8+5+8 6 = 7 Forecasting by last period’s sales This method eliminates the influence of old data and bases the forecast only upon the sales of the previous period. Example: Period No. Actual Sales Forecast sales Errors in forecast 1 5 2 4 5 +1 3 8 4 -4 4 7 8 +1 5 4 7 +3 4 Forecasting by Moving Average This is a compromise between the two methods explained above, in which the forecast is neither
  8. 8. 8 | influenced by very old data nor does it solely reflect the figure of the previous period. Consider the sales figures shown in the table below, which can be used to construct a sales forecast for the next year. Year Period Sales Four-period moving average forecast 1987 1 50 2 60 3 50 4 40 1988 1 50 50 2 55 50 3 40 48.75 4 30 46.25 1989 1 35 43.75 2 45 40 3 35 37.5 4 25 36.25 1990 1 35 35 2 45 35 3 35 35 4 30 36.25 Weighted Moving Average (WMA) In simple moving average, we have equal effects to each component of the moving average data base, in a weighted moving average we can place any weights on each element, provided that the sum of all the weights is equal to one. Suppose that in a four month period the best forecast is derived by using 40% of the actual sales for the most recent month, 30% of the two month ago, 20% of three months ago and 10% of four months ago. If actual sales were as follows; Month 1 Month 2 Month 3 Month 4 Month 5 100 90 105 95 ? The forecast for month 5 can be found out by; 𝐹𝑠 = 0.40(95) + 0.30(105) + 0.20(90) + 0.10(100) = 97.5 Let the actual sales for month 5 came out to be 110, then the month 6 forecast will be 𝐹𝑠 = 0.40(110) + 0.30(95) + 0.20(105) + 0.10(90) = 102.5 Forecasting by exponential smoothening With the help of this technique, we just need to retain the previous forecast figure and the latest actual sales figure. 𝑛𝑒𝑀 π‘“π‘œπ‘Ÿπ‘’π‘π‘Žπ‘ π‘‘ = 𝛼(π‘™π‘Žπ‘‘π‘’π‘ π‘‘ π‘ π‘Žπ‘™π‘’π‘  π‘“π‘–π‘”π‘’π‘Ÿπ‘’) + (1 βˆ’ 𝛼)(π‘œπ‘™π‘‘ π‘“π‘œπ‘Ÿπ‘’π‘π‘Žπ‘ π‘‘) The term β€˜Ξ±β€™ is known as smoothing constant The use of this technique permits to respond to recent actual events, but at the same time maintain certain amount of stability. The smoothing constant indicates the amount by which the new forecast responds to the latest sales figure, and its value lies between 0.1 to 0.3 To find out the smoothing constant, that gives the equivalent of an N-period moving average, use the relation, 𝛼 = 2 𝑁 + 1 Process planning It means the preparation of work detail plan, by determining the most economical method of performing an operation of activity. The information needed to do process planning is: Quantity of work to be done Quality of work to be completed Availability of tools, equipments and personnel Sequence of operations to be performed Standard time for each operation Procedure for process planning:- 1. Selection of process 2. Selection of material 3. Selection of jigs, fixtures and other special attachments 4. Selection of cutting tools and inspection gauges Operations Research & Inventory Control Introduction Let an industrialist has two industries (A and B) at different locations. He wants to send the finished goods to five different locations. To do this task, there are several ways. The point of discussion here is which way, out of the several ways, is the best alternative to send the goods to the five locations in sight. For the industrialist in question, the best alternative would be the one in which he has to pay minimum transportation charges, which becomes an Optimum condition. By the term optimum condition means, a point where all the conditions are favourable. The approach to optimisation involves the following;- The criteria which judges the best of the several alternatives Characteristics of the various alternatives being judged Methods available to judge the best performance for the selected criteria Operations Research It signifies research on operations by taking into consideration a particular view of operations and a particular kind of research. The purpose of this subject
  9. 9. 9 | and techniques used is to provide the management with explicit quantitative understanding and assessment of complex situations, helping them make better decisions. Linear Programming It is defined as the optimisation of a linear function of variables subject to constraints of linear inequalities. A LPP consists of three components: 1. Decision variables (activities) 2. The objective function (Goal) 3. The constraints (restrictions) Objective function: It is a clearly identifiable and measureable quantity Constraints: These are limited resources within which we have to obtain optimised solution for the objective function. Three different types of solution: 1. Infinite Solution: The objective function slope equals to one of the constraints which forms the boundary 2. No solution: These is no solution possible for the given LPP 3. Unbounded Solution: The greatest value of objective function occurs at infinity and it simply means that the common feasible region is not bounded by limits on constraints Simplex method Procedure RHS of each constraint should be non negative Each decision variable of the problem should be non negative Inequalities in constraints should be converted to equalities Set m = no. of equality constraints and n= no. of variables Put (n-m) variable equal to zero (n-m) = non basic variable m= basic variable Special case: Infinite solution: When a non basic variable in an optimal solution has a zero value for Ξ”j row then the solution is not unique Unbounded solution: When all replacement ratios are either infinite (or) negative then the solution terminates. This indicates the problem has unbounded solution Infeasible solution: When in the final solution an artificial variable is in the basis then there is no feasible solution to the problem Duality in LP For every LP problem there exists a related unique LP problem involving the same data which also describes and solves original problem Primal Dual Maximum Minimum No. of variables No. of constraints No. of constraints No. of variables type of constraints Non negative variables = type constraints Unrestricted variables Unrestricted variable = type constraints Big M method: In those situations where an identity matrix is not obtained initially another form of simplex method called Big M method is applied. In this method, artificial variable are put into the model to obtain an initial solution. Transportation Problem These problems are used for meeting the supply and demand requirements under given conditions in the best optimal effective manner. Cij= cost of transportation of one unit from the ith source to the jth destination Xij= Quantity to be transported from ith source to the jth destination π‘‘π‘œπ‘‘π‘Žπ‘™ π‘‘π‘Ÿπ‘Žπ‘›π‘ π‘π‘œπ‘Ÿπ‘‘π‘Žπ‘‘π‘–π‘œπ‘› π‘π‘œπ‘ π‘‘ = βˆ‘ βˆ‘ 𝐢𝑖𝑗 𝑋𝑖𝑗 π‘š 𝑗=1 𝑛 𝑖=1 Feasible Solution: A set of non negative individual allocations which also satisfy the given constraints Basic Feasible Solution: A basic feasible solution of mXn TP is basic feasible if the total number of allocations is exactly the equal to (m+n-1) Optimal Solution: A feasible solution is said to be optimal, if it minimizes the total transportation cost. Non degenerate Basic Feasible Solution: A feasible solution of mXn TP is non degenerate is Total number of allocations is exactly equal to (m+n-1) These allocations are in independent positions Unbalanced TP If total supply from all sources equals total demand in all destinations, then the TP is balanced, otherwise unbalanced If given TP is unbalanced, then make the problem balanced by using a dummy source or destination Degeneracy When the number of allocations are less then (m+n-1) then optimality test cannot be performed and such a solution is called degenerate solution. Assignment problem These problems are special cases of TP where Matrix must be a square matrix The optimal solution to the problem would always be such that there would only be one assignment in the given row or column Hungarian method is used to solve an AP Steps for applying Hungarian method: Subtract the smallest element of each row to the every element of corresponding row
  10. 10. 10 | Subtracting the smallest element of each column to every element in corresponding columns Now all zeroes are to be covered with minimum number of lines If number of lines is equal to the number of rows or columns then optimal solution is obtained Queuing Theory Queue means the number of customers waiting to be serviced. The queue does not include the customer being serviced. The process which serves the customer is called service facility. Elements of a Queuing system: 1. Input or arrival process: Size of queue Pattern of arrivals Customer’s behaviour 2. Queue discipline 3. Service mechanism Single queue one server Single queue several server Several queue one server Several queue several server 4. Capacity of the system Operating characteristics of a Queuing system: Expected number of customers in the system is denoted by [E(n)] or L. it is the average number of customers in the system, both waiting and being serviced Expected number of customers in the queue [E(m)] or Lq. it is the average number of customers waiting in queue Here, m=n-1, i.e. excluding the customer being serviced, or Lq = L-1 Expected waiting time in the system E(v) or w, is the average total time spent by a customer in the system. It is generally taken to be equal to waiting time + service time Expected waiting time in queue denoted by E(w) or wq. it is the average time spent by a customer in the queue before the commencement of the service The server utilization factor, 𝑃 = πœ†/πœ‡. It is the proportion of time that a service actually stands with a customer, Here πœ† = average number of customers arriving per unit time and πœ‡ = average number of customers completing service per unit time. The value P is also known as traffic intensity or the clearing ratio Deterministic queuing system A queuing system wherein the customers arrive at regular intervals and service time for each customer is known and constant. 1. if > πœ‡ , the waiting line shall be formed and will increase indefinitely. The service facility would always be busy and service system will eventually fail 2. if πœ† ≀ πœ‡, there shall be no queue and hence no waiting time. The proportion of time the service facility would be idle is 1 βˆ’ πœ† πœ‡ . Probabilistic Queuing system It is assumed that customers joining the queuing system arrive in a random manner (variable) and follow a Poisson’s distribution. Queuing Model N(t) = no. of customers in queuing system at time t S = no. of servers in queuing system Pn(t) = Probability of n units in queuing system πœ† 𝑛= mean arrival rate (units/ unit time) Lq = average number of customers in queue system n= mean number of units in the queuing system including the one being served ws= average waiting time in the queue wq= average time the queue system M/M/1 : (∞/ FIFO) Single service channel, Poisson’s input, exponential service, no limit on the system capacity. First In First Out. 𝑃𝑛 = 𝜌 𝑛(1 βˆ’ 𝜌), π‘€β„Žπ‘’π‘Ÿπ‘’ 𝜌 = πœ† πœ‡ < 1 π‘Žπ‘›π‘‘ 𝑛 β‰₯ 0 inter arrival time = 1/Ξ» traffic intensity facto, 𝜌 = πœ†/πœ‡ average number of customers in system, 𝐿 = 𝜌/(1 βˆ’ 𝜌) average number of customer of queue or average queue length, 𝐿 π‘ž = 𝐿 βˆ’ 𝜌 Waiting time in system, 𝑀 = 1 πœ‡βˆ’πœ† Average or expected waiting time in queue, π‘€π‘ž = 𝐿 π‘ž πœ† Probability that the service facility is idle, 𝑃0(𝑑) = (1 βˆ’ 𝜌) Probability that the service facility has n customers at time t, 𝑃𝑛(𝑑) = 𝜌 𝑛 𝑃0(𝑑) Average length of non empty queue, 𝐿 𝑛 = 1 1βˆ’πœŒ The fluctuation of queue length, 𝑣(𝑛) = 𝜌 (1βˆ’πœŒ)2 Probability of n arrivals in time t, 𝑃(𝑛𝑑) = π‘’βˆ’πœ†(πœ†π‘‘) 𝑛 𝑛! Probability that the waiting time in the queue is greater than or equal to t, 𝑃(π‘€π‘ž β‰₯ 𝑑) = πœ† π‘’βˆ’(πœ‡βˆ’πœ†)𝑑 πœ‡ Probability that waiting time in system is greater than or equal to t, 𝑃 = π‘’βˆ’(πœ‡βˆ’πœ†)𝑑 Probability that waiting time in system is less than or equal to t, 𝑃 = 1 βˆ’ π‘’βˆ’(πœ‡βˆ’πœ†)𝑑 Inventory Control Inventory is defined as the list of movable goods which helps directly or indirectly in production of goods for sale. We can also defined inventory as a comprehensive of goods for sale. We can also defined inventory as a comprehensive list of movable items which are required
  11. 11. 11 | for manufacturing the products and to maintain the plant facilities in working conditions. It can be divided in two parts. Direct Inventories The inventories which play a direct role in manufacturing of a product and become an integral part of the finished product are called direct inventories. e.g., raw materials, purchases part and finished goods. Indirect inventories The inventories which helps the raw material to get converted into products but not integral part of finished product is called indirect inventories, e.g., tool and supplies (material used in running the plant but do not go into the product) are indirect inventories. Inventory Control It means making the desired item of required quality and in required quantity available to various departments when needed. Determining Inventory Control The amount of inventory, a company should carry is determined by five basic variables. (a) Order quantity (b) Reorder point (c) Lead time (d) Safety stock (e) Butter stock Order Quantity It is the volume of stock at which order is placed or total quantity of buy or sell order. Reorder Point It is time between initiating the order and receiving the required quantity. Reorder point = Minimum inventory + Procurement time Consumption rate. Lead Time The time gap between placing of an order an its actual arrival in the inventory is known as lead time. It consist of requisition time and procurement time. It has two components. Administrative Lead Time From initiation of procurement action until the placing of an order. Delivery Lead Time From placing of an order until the delivery of the odered material. Safety Stock If the maximum inventory would be equal to the order quantity Q and minimum inventory would be zero. Average inventory in this case = Q 2 Safety stock = k Average Consumption duringlead time k = A factor abased on acceptable frequency of stock out in a given number of years. Buffer Stock For an average demand during average lead time the additional stock termed as buffer stock. Buffer stock = Average demand Average lead time When no stock outs are desired Buffer stock = Maximum demand during lead time (DDLT) Average Demand During Lead Time (DDLT) When demand rate varies about the average demand during a constant lead time (LT) period Reorder Level (ROL) = Average (DDLT) LT + BS Inventory Cost The costs that are affected by firm’s decision to maintain particular level of inventory are called cost associated with inventories or relevant inventory cost. Total Inventory Costs (TIC) TIC = Purchase cost + Total Variable Cost (TVC) of managing the inventory TIC = Purchase cost + Inventory cost + Ordering cost + Shortage cost Purchase Cost It is defined as the cost of purchasing a unit of an item. Purchase cost = Price per unit Demand per unit time where, Cu = Unit cost D = Annual demand Ordering Cost It is defined as the cost of placing an order from a vector. This represents the expenses involved in placing an order with the outside supplier. This includes the costs involved in processing and ordering for purchase. expediting over the orders, receiving the consignment and inspection.
  12. 12. 12 | Annual ordering cost= oC Q D where, Q = Produced purchased or supplied throughout the entire time period (one year) or order quantity. Co = Cost of placing an order D = Annual demand Carrying Cost Carrying or holding costs are the costs incurred maintaining the stores in the firm. It is proportional to the amount of inventory and the time over which it is held. Annual carrying cost Cui = Cu i a 2 where, Cu = Unit purchase cost, i = Interest rate Shortage Cost When an item cannot be supplied on consumer’s demand, the penalty cost for running out of stock is called shortage cost or stock out cost. Shortage cost = Cost of being short one unit in inventory Average number of unit short in the inventory. Economic Order Quantity (EOQ) Economic order quantity is the order quantity that minimizes total inventory holding costs and ordering costs. It is one of the oldest classical production scheduling models. Assumptions of EOQ Model The ordering cost is constant The rate of demand is known Lead time is fixed Purchase price of the item is constant Replenishment is made Only one product is involved. instantaneously EOQ When Stock Replenishment is Instantaneous Economic Order Quantity (EOQ) is the order quantity that minimizes total inventory holding costs and ordering costs. It is one of the oldest classical production scheduling models. It is defined as the quantity which will minimise the total variable cost of managing the inventory. TVC = oC Q D Q 2 Cu i EOQ (Q ) = 0 u 2C D C i where, Co = Cost of placing an order, Cu = Unit purchase cost i = interest rate, D= Annual consumption of the product. Optimum number of order placed per year no= o D Q = u o DC i 2C where, Qo = Economic order quantity Inventory Models Broadly categorising, inventory models are of two type: 1. Static Inventory Model Only one order can be placed to meet the demand, as repeated orders are deemed too expensive 2. Dynamic Inventory Model In this model, the order can be repeated again and again to replenish the stock. Further, the dynamic inventory model can be classified into: a) Deterministic Model Both the lead time and demand for an item are pre determined b) Probabilistic Model In this, both the demand and lead time are not known as both keep on varying Questions: 1. Fifty observations of a production operation revealed a mean cycle time of 10 min . The worker was evaluated to be performing at 90% efficiency. Assuming the allowances to be 10% of the normal time, the standard time (in second) for thejob is (A) 0.198 (B) 7.3 (C) 9.0 (D) 9.9 2. When using a simple moving average to forecast demand, one would (A) give equal weight to all demand data (B) assign more weight to the recent demand data (C) include new demand data in the average without discarding the earlier data (D) include new demand data in the average after discarding some of the earlier demand data 3. Production flow analysis (PFA) is a method of identifying part families that uses data from (A) engineering drawings (B) production schedule (C) bill of materials (D) route sheets 4. A project consists ofthree parallel paths with mean durations and variances of (10, 4) , (12, 4) and (12, 9) respectively. According to the standard PERT assumptions, the distribution of the project duration is (A) beta with mean 10 and standard deviation 2 (B) beta with mean 12 and standard deviation 2 (C) normal with mean 10 and standard deviation 3 (D) normal with mean 12 and standard deviation 3 5. The supplies at three sources are 50, 40 and 60 unit respectively whilst the demands at the four
  13. 13. 13 | destinations are 20, 30, 10 and 50 unit. In solving this transportation problem (A) a dummy source of capacity 40 unit is needed (B) a dummy destination of capacity 40 unit is needed (C) no solution exists as the problem is infeasible (D) no solution exists as the problem is degenerate 6. Arrivals at a telephone booth are considered to be Poisson, with an average time of 10 minutes between successive arrivals. The length of a phone call is distributed exponentially with mean 3 minutes. The probability that an arrival does not have to wait before service is (A) 0.3 (B) 0.5 (C) 0.7 (D) 0.9 7. An item can be purchased for Rs. 100. The ordering cost is Rs. 200 and the inventory carrying cost is 10% of the item cost 𝑝 er annum. If the annual demand is 4000 unit, the economic order quantity (in unit) is (A) 50 (B) 100 (C) 200 (D) 400 8. In carrying out a work sampling study in a machine shop, it was found that a particular lathe was down for 20% of the time. What would be the 95% confidence interval of this estimate, if 100 observations were made? (A) (0.16, 0.24) (B) (0.12, 0.28) (C) (0.08, 0.32) (D) None of these 9. The standard time of an operation while conducting a time study is (A) mean observed time + allowances (B) normal time + allowances (C) mean observed time Γ— rating factor + allowances (D) normal time Γ— rating factor + allowances 10. The principles of motion economy are mostly used while conducting (A) a method study on an operation (B) a time study on an operation (C) a financial appraisal of an operation (D) a feasibility study of the proposed manufacturing plant 11. A project consists of activities 𝐴 to 𝑀 shown in the net in the following figure with the duration of the activities marked in days The project can be completed (A) between 18, 19 days (B) between 20, 22 days (C) between 24, 26 days (D) between 60, 70 days 12. A manufacturer produces two types ofproducts, 1 and 2, at production levels of π‘₯1 and π‘₯2 respectively. The profit is given is 2π‘₯1 + 5π‘₯2. The production constraints are π‘₯1 + 3π‘₯2 ≀ 40 3π‘₯1 + π‘₯2 ≀ 24 π‘₯1 + π‘₯2 ≀ 10 π‘₯1 > 0, π‘₯2 > 0 The maximum profit which can meet the constraints is (A) 29 (B) 38 (C) 44 (D) 75 13. Market demand for springs is 8,00,000 per annum. A company purchases these springs in lots and sells them. The cost of making a purchase order is Rs. 1200. The cost of storage of springs is Rs. 120 per stored piece per annum. The economic order quantity is (A) 400 (B) 2,828 (C) 4,000 (D) 8,000 14. The sale of cycles in a shop in four consecutive months are given as 70, 68, 82, 95. Exponentially smoothing average method with a smoothing factor of 0.4 is used in forecasting. The expected number of sales in the next month is (A) 59 (B) 72 (C) 86 (D) 136 15. A residential school stipulates the study hours as 8.00 pm to 10.30 π‘π‘š. Warden makes random checks on a certain student 11 occasions a day during the study hours over a period of10 days and observes that he is studying on 71 occasions. Using 95% confidence interval, the estimated minimum hours of his study during that 10 day period is (A) 8.5 hours (B) 13.9 hours (C) 16.1 hours (D) 18.4 hours 16. Two machines of the same production rate are available for use. On machine 1, the fixed cost is Rs. 100 and the variable cost is Rs. 2 per piece produced. The corresponding numbers for the machine 2 are Rs. 200 and 𝑅𝑒. 1 respectively. For certain strategic reasons both the machines are to be used concurrently. The sales price of the first 800 units is Rs. 3.50 per unit and subsequently it is only Rs. 3.00. The breakeven production rate for each machine is (A) 75 (B) 100 (C) 150 (D) 600 17. The symbol used for Transport in work study is (A)β‡’ (B) 𝑇 (C) (D) 𝛻 18. A company produces two types of toys: 𝑃 and 𝑄. Production time of 𝑄 is twice that of 𝑃 and the company has a maximum of2000 time units per day. The supply of raw material is just sufficient to produce 1500 toys (of any type) per day. Toy type 𝑄 requires an electric switch which is available @ 600 pieces per day only. The company makes a profit of
  14. 14. 14 | Rs. 3 and Rs. 5 on type 𝑃 and 𝑄 respectively. For maximization ofprofits, the daily production quantities of 𝑃 and 𝑄 toys should respectively be (A) 1000, 500 (B) 500, 1000 (C) 800, 600 (D) 1000, 1000 19. A company has an annual demand of 1000 units, ordering cost of Rs. 100 / order and carrying cost of Rs. 100/ unit/year. If the stock‐out cost are estimated to be nearly Rs. 400 each 𝑑 ime the company runs out‐of‐stock, then safety stock justified by the carrying cost will be (A) 4 (B) 20 (C) 40 (D) 100 20. A maintenance service facility has Poisson arrival rates, negative exponential service time and operates on a β€˜first come first served’ queue discipline. Breakdowns occur on an average of 3 per day with a range of zero to eight. The maintenance crew can service an average of 6 machines per day with a range of zero to seven. The mean waiting time for an item to be serviced would be (A) 1 6 day (B) 1 3 day (C) 1 day (D) 3 day 21. An electronic equipment manufacturer has decided to add a component sub‐ assembly operation that can produce 80 units during a regular 8‐hours shift. This operation consist of three activities as below Activity Standard time ( min ) M. Mechanical assembly 12 E. Electric wiring 16 T. Test 3 For line balancing the number of work stations required for the activities 𝑀, 𝐸 and 𝑇 would respectively be (A) 2, 3, 1 (B) 3, 2, 1 (C) 2, 4, 2 (D) 2, 1, 3 22. A soldering operation was work‐sampled over two days (16 hours) during which an employee soldered 108 joints. Actual working time was 90% of the total time and the performance rating was estimated to be 120 per cent. If the contract provides allowance of 20 percent of the time available, the standard time for the operation would be (A) 8 min (B) 8.9 min (C) 10 min (D) 12 min 23. A standard machine tool and an automatic machine tool are being compared for the production of component. Following data refers to the two machines. Standard Machine Tool Automatic Machine Tool Setup time 30 min 2 hours Machining time per piece 22 min 5 min Machine rate Rs. 200 per hour Rs. 800 per hour The break even production batch size above which the automatic machine tool will be economical to use, will be (A) 4 (B) 5 (C) 24 (D) 225 24. There are two products 𝑃 and 𝑄 with the following characteristics Product Deman d (Units) Order cost ( 𝑅𝑠/ order) Holding Cost (Rs. /unit/year) 𝑃 100 50 4 𝑄 400 50 1 The economic order quantity (EOQ) of products 𝑃 and 𝑄 will be in the ratio (A) 1 : 1 (B) 1 : 2 (C) 1 : 4 (D) 1 : 8 25. For a product, the forecast and the actual sales for December 2002 were 25 and 20 respectively. If the exponential smoothing constant (𝛼) is taken as 0.2, then forecast sales for January 2003 would be (A) 21 (B) 23 (C) 24 (D) 27 26. In PERT analysis a critical activity has (A) maximum Float (B) zero Float (C) maximum Cost (D) minimum Cost Common Data for 𝑄.27 and 𝑄.28 Consider a linear programming problem with two variables and two constraints. The objective function is: Maximize𝑋1 + 𝑋2. The corner points of the feasible region are (0,0), (0,2), (2,0) and (4/3, 4/3) 27. If an additional constraint 𝑋1 + 𝑋2 ≀ 5 is added, the optimal solution is (A) ( 55 3’3 ) (𝐡) ( 4 3’ 4 3 ) (C) ( 55 2’2 ) (D) (5, 0) 28. Let π‘Œ1 and π‘Œ2 be the decision variables of the dual and 𝑣1 and 𝑣2 be the slack variables of the dual of the given linear programming problem. The optimum dual variables are (A) π‘Œ1 and π‘Œ2 (B) π‘Œ1 and 𝑣1 (C) π‘Œ1 and 𝑣2 (D) 𝑣1 and 𝑣2 29. A company has two factories 1, 𝑆2, and two warehouses 𝐷1, 𝐷2. The supplies from 𝑆1 and 𝑆2 are 50 and 40 units respectively. Warehouse 𝐷1 requires a minimum of 20 units and a maximum of 40 units. Warehouse 𝐷2 requires a minimum of 20 units and, over and above, it can take as much as can be supplied. A balanced transportation problem is to be formulated for the above situation. The number of supply points, the number of demand points, and the total supply (or total demand) in the balanced transportation problem respectively are (A)2,4,90 (𝐡)2,4,110
  15. 15. 15 | (C)3,4,90 (𝐷)3,4,110 30. A project has six activities (A to 𝐹) with respective activity duration 7, 5, 6, 6, 8, 4 days. The network has three paths A‐B, C‐D and E‐F. All the activities can be crashed with the same crash cost per day. The number of activities that need to be crashed to reduce the project duration by 1 day is (A) 1 (B) 2 (C) 3 (D) 6 31. The distribution of lead time demand for an item is as follows: Lead time demand 𝑃 robability 80 0.20 100 0.25 120 0.30 140 0.25 The reorder level is 1.25 times the expected value of the lead time demand. The service level is (A) 25% (B) 50% (C) 75% (D) 100% 32. A welding operation is time‐studied during which an operator was pace‐rated as 120%. The operator took, on an average, 8 minutes for producing the weld‐ joint. If a total of10% allowances are allowed for this operation. The expected standard production rate of the weld‐joint (in units per 8 hour day) is (A) 45 (B) 50 (C) 55 (D) 60 33. A component can be produced by any of the four processes I, II, III and IV. Process I has a fixed cost of Rs. 20 and variable cost of Rs. 3 per piece. Process II has a fixed cost Rs. 50 and variable cost of Rs. 1 per piece. Process III has a fixed cost of Rs. 40 and variable cost of Rs. 2 per piece. Process IV has a fixed cost of Rs. 10 and variable cost of Rs. 4 per piece. If the company wishes to produce 100 pieces of the component, form economic point of view it should choose (A) Process I (B) Process II (C) Process III (D) Process IV 34. Consider a single server queuing model with Poisson arrivals (πœ† = 4/hour) and exponential service (πœ‡ = 4/hour) . The number in the system is restricted to a maximum of 10. The probability that a person who comes in leaves without joining the queue is (A) 1 11 (B) 1 10 (C) 1 9 (D) 1 2 35. The sales of a product during the last four years were 860, 880, 870 and 890 units. The forecast for the fourth year was 876 units. If the forecast for the fifth year, using simple exponential smoothing, is equal to the forecast using a three period moving average, the value ofthe exponential smoothing constant 𝛼 is (A) 1 7 (B) 1 5 (C) 2 7 (D) 2 5 36. An assembly activity is represented on an Operation Process Chart by the symbol (A) β–‘ (B) 𝐴 (C) 𝐷 (D) 𝑂 37. The standard deviation of the critical path of the project is (A)√151 days (𝐡)√155 days (C)√200 days (𝐷)√238 days 38. The expected completion time of the project is (A) 238 days (B) 224 days (C) 171 days (D) 155 days 39. The table gives details of an assembly line. Work station I II III IV V VI Totaltask time at the workstation (in minutes) 7 9 7 10 9 6 What is the line efficiency of the assembly line? (A) 70% (B) 75% (C) 80% (D) 85% 40. A stockist wishes to optimize the number of perishable items he needs to stock in any month in his store. The demand distribution for this perishable item is Demand (in unit s) 2 3 4 5 𝑃 robability 0.10 0.35 0.35 0.20 The stockist pays Rs. 70 for each item and he sells each at Rs. 90. If the stock is left unsold in any month, he can sell the item at Rs. 50 each. There is no penalty for unffilfilled demand. To maximize the expected profit, the optimal stock level is (A) 5 units (B) 4 units (C) 3 unit 𝑠 (D) 2 unit 𝑠 41. Consider the following data for an item. Annual demand : 2500 units per year, Ordering cost : Rs. 100 𝑝 er order, Inventory holding rate : 25% of unit price Price quoted by a supplier Order quantity (units) Unit price (Rs.) < 500 10 β‰₯ 500 9 The optimum order quantity (in units) is (A) 447 (B) 471 (C) 500 (D) β‰₯ 600 42. An manufacturing shop processes sheet metal jobs, wherein each job must pass through two machines (𝑀𝑙 π‘Žπ‘›π‘‘ 𝑀2, 𝑖𝑛 that order). The processing time (in hours) for these jobs is
  16. 16. 16 | Machine Jobs P Q R S T U M1 15 32 8 27 11 16 M2 6 19 13 20 14 7 The optimal make‐span (in‐hours) of the shop is (A) 120 (B) 115 (C) 109 (D) 79 43. A firm is required to procure three items (𝑃, 𝑄, π‘Žπ‘›π‘‘ 𝑅) . The prices quoted for these items (in Rs.) by suppliers 𝑆1, 𝑆2 and 𝑆3 are given in table. The management policy requires that each item has to be supplied by only one supplier and one supplier supply only one item. The minimum total cost (in Rs.) of procurement to the firm is Item Suppliers S1 S2 S3 P 110 120 130 Q 115 140 140 R 125 145 165 (A) 350 (C) 385 (B) 360 (D) 395 44. In an MRP system, component demand is (A) forecasted (B) established by the master production schedule (C) calculated by the MRP system from the master production schedule (D) ignored 45. The number of customers arriving at a railway reservation counter is Poisson distributed with an arrival rate of eight customers per hour. The reservation clerk at this counter takes six minutes per customer on an average with an exponentially distributed service time. The average number of the customers in the queue will be (A) 3 (B) 3.2 (C) 4 (D) 4.2 46. The net requirements of an item over 5 consecutive weeks are 50‐0‐15‐20‐20. The inventory carrying cost and ordering cost are Rs. 1 per item per week and Rs. 100 per order respectively. Starting inventory is zero. Use β€œLeast Unit Cost Technique” for developing the plan. The cost of the plan (in Rs.) is (A) 200 (B) 250 (C) 225 (D) 260 47. In a machine shop, pins of 15 mm diameter are produced at a rate of 1000 per month and the same is consumed at a rate of 500 per month. The production and consumption continue simultaneously till the maximum inventory is reached. Then inventory is allowed to reduced to zero due to consumption The lot size of production is 1000. If backlog is not allowed, the maximum inventory level is (A) 400 (B) 500 (C) 600 (D) 700 48. The maximum level of inventory of an item is 100 and it is achieved with infinite replenishment rate. The inventory becomes zero over one and half month due to consumption at a uniform rate. This cycle continues throughout the year. Ordering cost is Rs. 100 per order and inventory carrying cost is Rs. 10 per item per month. Annual cost (in Rs.) of the plan, neglecting material cost, is (A) 800 (B) 2800 (C) 4800 (D) 6800 49. Capacities of production of an item over 3 consecutive months in regular time are 100, 100 and 80 and in overtime are 20, 20 and 40. The demands over those 3 months are 90, 130 and 110. The cost of production in regular time and overtime are respectively Rs. 20 𝑝 er it em and Rs. 24 𝑝er item. Inventory carrying cost is Rs. 2 per item per month. The levels of starting and final inventory are nil. Backorder is not permitted. or minimum cost of plan, the level of planned production in overtime in the third month is (A) 40 (B) 30 (C) 20 (D) 0 Common Data For 𝑄.50 and 𝑄.51 Consider the Linear Programme (𝐿𝑃) Max 4π‘₯ + 6𝑦 Subject to 3π‘₯ + 2𝑦 ≀ 6 2π‘₯ + 3𝑦 ≀ 6 π‘₯, 𝑦 β‰₯ 0 50. After introducing slack variables 𝑠 and 𝑑, the initial basic feasible solution is represented by the table below (basic variables are 𝑠 = 6 and 𝑑 = 6, and the objective function value is 0) βˆ’4 βˆ’6 0 0 0 𝑠 3 2 1 0 6 𝑑 2 3 0 1 6 π‘₯ 𝑦 𝑠 𝑑 RHS After some simplex iterations, the following table is obtained 0 0 0 2 12 𝑠 5/3 0 1 βˆ’1/3 2 𝑦 2/3 1 0 1/3 2 π‘₯ 𝑦 𝑠 𝑑 RHS From this, one can conclude that (A) the 𝐿𝑃 has a unique optimal solution (B) the 𝐿𝑃 has an optimal solution that is not unique (C) the 𝐿𝑃 is infeasible (D) the 𝐿𝑃 is unbounded 51. The dual for the 𝐿𝑃 in 𝑄. 50 is (A) Min 6𝑒 + 6𝑣 (B) Max 6𝑒 + 6𝑣 subject to 3𝑒 + 2𝑣 β‰₯ 4
  17. 17. 17 | 2𝑒 + 3𝑣 β‰₯ 6 𝑒, 𝑣 β‰₯ 0 (C) Max 4𝑒 + 6𝑣 subject to 3𝑒 + 2𝑣 β‰₯ 6 2𝑒 + 3𝑣 β‰₯ 6 𝑒, 𝑣 β‰₯ 0 subject to 3𝑒 + 2𝑣 ≀ 4 2𝑒 + 3𝑣 ≀ 6 𝑒, 𝑣 β‰₯ 0 (D) Min 4𝑒 + 6𝑣 subject to 3𝑒 + 2𝑣 ≀ 6 2𝑒 + 3𝑣 ≀ 6 𝑒, 𝑣 β‰₯ 0 52. The product structure of an assembly 𝑃 is shown in the figure. Estimated demand for end product 𝑃 is as follows Week 1 2 3 4 5 6 Demand 1000 1000 1000 1000 1200 1200 ignore lead times for assembly and sub‐assembly. Production capacity (per week) for component 𝑅 is the bottleneck operation. Starting with zero inventory, the smallest capacity that will ensure a feasible production plan up to week 6 is (A) 1000 (B) 1200 (C) 2200 (D) 2400 53. For the network below, the objective is to find the length of the shortest path from node 𝑃 to node 𝐺. Let 𝑑𝑖 𝑗 . be the length of directed arc from node 𝑖 to node 𝑗. Let 𝑆𝑗 be the length of the shortest path ffom 𝑃 to node 𝑗. Which of the following equations can be used to find 𝑆 𝐺 ? (A) 𝑆 𝐺 = Min {𝑆 𝑄, 𝑆 𝑅} (B) (𝐡) 𝑆 𝐺 = Min {𝑆 𝑄 βˆ’ 𝑑 𝑄𝐺, 𝑆 𝑅 βˆ’ 𝑑 𝑅𝐺} (C) 𝑆 𝐺 = Min {𝑆 𝑄 + 𝑑 𝑄𝐺, 𝑆 𝑅 + 𝑑 𝑅𝐺} (D) (𝐷) 𝑆 𝐺 = Min {𝑑 𝑄𝐺, 𝑑 𝑅𝐺} 54. A moving average system is used for forecasting weekly demand 𝐹1(𝑑) and 𝐹2(𝑑) are sequences of forecasts with parameters π‘š1 and π‘š2, respectively, where π‘š1 and π‘š2(π‘š1 > π‘š2) denote the numbers of weeks over which the moving averages are taken. The actual demand shows a step increase from 𝑑1 to 𝑑2 at a certain time. Subsequently, (A) neither 𝐹1(𝑑) nor 𝐹2(𝑑) will catch up with the value 𝑑2 (B) both sequences 𝐹1(𝑑) and 𝐹2(𝑑) will reach 𝑑2 in the same period (C) 𝐹1(𝑑) will attain the value 𝑑2 before 𝐹2(𝑑) (D) 𝐹2(𝑑) will attain the value 𝑑2 before 𝐹1(𝑑) 55. For the standard transportation linear programme with π‘š source and 𝑛 destinations and total supply equaling total demand, an optimal solution (lowest cost) with the smallest number of non‐zero π‘₯𝑖𝑗 values (amounts ffom source 𝑖 to destination j) is desired. The best upper bound for this number is (A) π‘šπ‘› (B) 2 (π‘š + 𝑛) (C) π‘š + 𝑛 (D) π‘š + 𝑛 βˆ’ 1 56. A set of 5 jobs is to be processed on a single machine. The processing time (in days) is given in the table below. The holding cost for each job is Rs. 𝐾 per day. Job Processing time 𝑃 5 𝑄 2 𝑅 3 𝑆 2 𝑇 1 A schedule that minimizes the total inventory cost is (A) T‐S‐Q‐R‐P (B) P‐R‐S‐Q‐T (C) T‐R‐S‐Q‐P (D) P‐Q‐R‐S‐T 57. In an 𝑀/𝑀/1 queuing system, the number ofarrivals in an interval oflength 𝑇 is a Poisson random variable (i.e. the probability of there being arrivals in an interval of length 𝑇 is π‘’βˆ’πœ†π‘‡(πœ†π‘‡) 𝑛 𝑛! ). The probability density function 𝑓(𝑑) of the inter‐arrival time is (A) πœ†2 (π‘’βˆ’πœ†2 𝑑 ) (B) π‘’βˆ’πœ†2 𝑑 πœ†2 (C) πœ†π‘’βˆ’πœ†π‘‘ (𝐷) π‘’βˆ’πœ†π‘‘ πœ† Common Data For 𝑄.58 and 𝑄.59 Consider the following PERT network: The optimistic time, most likely time and pessimistic time ofall the activities are given in the table below: Activity Optimisti c time (days) Most likely time (days) Pessimisti c time (days) 1 βˆ’ 2 1 2 3 1 βˆ’ 3 5 6 7 1 βˆ’ 4 3 5 7 2 βˆ’ 5 5 7 9 3 βˆ’ 5 2 4 6 5 βˆ’ 6 4 5 6 4 βˆ’ 7 4 6 8 6 βˆ’ 7 2 3 4
  18. 18. 18 | 58. The critical path duration of the network (in days) is (A) 11 (B) 14 (C) 17 (D) 18 59. The standard deviation of the critical path is (A) 0.33 (B) 0.55 (C) 0.77 (D) 1.66 60. Six jobs arrived in a sequence as given below: Jobs Processing Time (days) I 4 II 9 III 5 IV 10 V 6 VI 8 Average flow time (in days) for the above jobs using Shortest Processing time rule is (A) 20.83 (B) 23.16 (C) 125.00 (D) 139.00 61. Consider the following Linear Programming Problem (LPP): Maximize 𝑍 = 3π‘₯1 + 2π‘₯2 Subject to π‘₯1 ≀ 4 π‘₯2 ≀ 6 3π‘₯1 + 2π‘₯2 ≀ 18 π‘₯1 β‰₯ 0, π‘₯2 β‰₯ 0 (A) The LPP has a unique optimal solution (B) The 𝐿𝑃𝑃 is infeasible. (C) The 𝐿𝑃𝑃 is unbounded. (D) The LPP has multiple optimal solutions. 62. A company uses 2555 units of an item annually. Delivery lead time is 8 days. The reorder point (in number of units) to achieve optimum inventory is (A) 7 (B) 8 (C) 56 (D) 60 63. Which ofthe following forecasting methods takes a fraction of forecast error into account for the next period forecast? (A) simple average method (B) moving average method (C) weighted moving average method (D) exponential smoothening method 64. The expected time (𝑑 𝑒) of a PERT activity in terms of optimistic time 𝑑0, pessimistic time (𝑑 𝑝) and most likely time (𝑑𝑙) is given by (A) 𝑑 𝑒 = 𝑑 π‘œ+4𝑑 𝑇+𝑑 𝑝 6 (C) 𝑑 𝑒 = 𝑑 π‘œ+4𝑑 𝑇+𝑑 𝑝 3 (B) 𝑑 𝑒 = 𝑑 π‘œ+4𝑑 𝑝+𝑑 𝑇 6 (D) 𝑑 𝑒 = 𝑑 π‘œ+4𝑑 𝑝+𝑑 𝑇 3 Common Data For 𝑄.65 and 𝑄.66 Four jobs are to be processed on a machine as per data listed in the table. Job Processing time (in days) Due date 1 4 6 2 7 9 3 2 19 4 8 17 65. If the Earliest Due Date (EDD) rule is used to sequence the jobs, the number of jobs delayed is (A) 1 (B) 2 (C) 3 (D) 4 66. Using the Shortest Processing Time (𝑆𝑃𝑇) rule, total tardiness is (A) 0 (B) 2 (C) 6 (D) 8 67. The project activities, precedence relationships and durations are described in the table. The critical path of the project is Activity Precedence Duration (in days) 𝑃 - 3 𝑄 - 4 𝑅 𝑃 5 𝑆 𝑄 5 𝑇 𝑅, 𝑆 7 π‘ˆ 𝑅, 𝑆 5 𝑉 𝑇 2 π‘Š π‘ˆ 10 (A) P‐R‐T‐V (C) P‐R‐U‐W (B) Q‐S‐T‐V (D) Q‐S‐U‐W 68. Annual demand for window frames is 10000. Each frame cost Rs. 200 and ordering cost is Rs. 300 per order. Inventory holding cost is Rs. 40 per frame per year. The supplier is willing of offer 2% discount if the order quantity is 1000 or more, and 4% if order quantity is 2000 or more. If the total cost is to be minimized, the retailer should (A) order 200 frames every time (B) accept 2% discount (C) accept 4% discount (D) order Economic Order Quantity 69. Simplex method of solving linear programming problem uses (A) all the points in the feasible region (B) only the corner points of the feasible region (C) intermediate points within the infeasible region (D) only the interior points in the feasible region 70. Vehicle manufacturing assembly line is an example of (A) product layout (B) process layout (C) manual layout (D) fixed layout 71. Little’s law is a relationship between (A) stock level and lead time in an inventory system (B) waiting time and length of the queue in a queuing system
  19. 19. 19 | (C) number of machines and job due dates in a scheduling problem (D) uncertainty in the activity time and project completion time 72. The demand and forecast for February are 12000 and 10275, respectively. Using single exponential smoothening method (smoothening coefficient = 0.25) , forecast for the month of March is (A) 431 (B) 9587 (C) 10706 (D) 11000 Common Data For 𝑄.73 and 𝑄.74 One unit of product 𝑃1 requires 3 kg ofresources 𝑅1 and 1 kg ofresources 𝑅2 One unit of product 𝑃2 requires 2 kg of resources 𝑅1 and 2 kg of resources𝑅2. The profits per unit by selling product 𝑃1 and 𝑃2 are Rs. 2000 and Rs. 3000 respectively. The manufacturer has 90 kg of resources 𝑅1 and 100 kg of resources 𝑅2. 73. The unit worth of resources 𝑅2, i.e., dual price of resources 𝑅2 in Rs. per kg is (A) 0 (B) 1350 (C) 1500 (D) 2000 74. The manufacturer can make a maximum profit of Rs. (A) 60000 (B) 135000 (C) 150000 (D) 200000 75. The word β€˜kanban’ is most appropriately associated with (A) economic order quantity (B) just‐in‐time production (C) capacity planning (D) product design 76. Cars arrive at a service station according to Poisson’s distribution with a mean rate of 5 per hour. The service time per car is exponential with a mean of 10 minutes. At steady state, the average waiting time in the queue is (A) 10 minutes (B) 20 minutes (C) 25 minutes (D) 50 minutes Common Data For 𝑄.77 and 𝑄.78 For a particular project, eight activities are to be carried out. Their relationships with other activities and expected durations are mentioned in the table below. Activit y Predecess ors Durations (days) π‘Ž 3 𝑏 π‘Ž 4 𝑐 π‘Ž 5 𝑑 π‘Ž 4 𝑒 𝑏 2 𝑓 𝑑 9 𝑔 𝑐, 𝑒 6 β„Ž 𝑓, 𝑔 2 77. The critical path for the project is (A) π‘Ž βˆ’ 𝑏 βˆ’ 𝑒 βˆ’ 𝑔 βˆ’ β„Ž (B) π‘Ž βˆ’ 𝑐 βˆ’ 𝑔 βˆ’ β„Ž (C) π‘Ž βˆ’ 𝑑 βˆ’ 𝑓 βˆ’ β„Ž (D) π‘Ž βˆ’ 𝑏 βˆ’ 𝑐 βˆ’ 𝑓 βˆ’ β„Ž 78. If the duration of activity 𝑓 alone is changed from 9 to 10 days, then the (A) critical path remains the same and the total duration to complete the project changes to 19 days. (B) critical path and the total duration to complete the project remains the same. (C) critical path changes but the total duration to complete the project remains the same. (D) critical path changes and the total duration to complete the project changes to 17 days. 79. Which one of the following is NOT a decision taken during the aggregate production planning stage? (A) Scheduling of machines (B) Amount oflabour to be committed (C) Rate at which production should happen (D) Inventory to be carried forward ANSWERS: 1 D 31 D 61 D 2 D 32 A 62 C 3 D 33 B 63 D 4 B 34 A 64 A 5 B 35 C 65 C 6 C 36 D 66 D 7 D 37 A 67 D 8 B 38 D 68 C 9 B 39 C 69 D 10 A 40 A 70 A 11 C 41 C 71 B 12 A 42 B 72 C 13 C 43 C 73 A 14 B 44 C 74 B 15 C 45 B 75 B 16 A 46 B 76 D 17 A 47 B 77 C 18 A 48 D 78 A 19 C 49 B 79 A 20 A 50 B 21 A 51 A 22 D 52 C 23 D 53 C 24 C 54 D 25 C 55 D 26 B 56 A 27 B 57 C 28 D 58 D 29 C 59 C 30 C 60 A